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An inductor $(L=100 \mathrm{mH})$, a resistor $(R=100 \Omega)$ and a battery $(E=100 \mathrm{~V})$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $A$ and $B$. The current in the circuit $1 \mathrm{~mm}$ after the circuit is
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$1 / \mathrm{e} \mathrm{A}$
$1 / \mathrm{e} \mathrm{A}$
$\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}=\frac{1}{\mathrm{e}} \mathrm{A}$
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